PRACTICE PROBLEMS

Sample Problem

Calculate the enthalpy change for the reaction in which hydrogen gas, H2 (g), is combined with fluorine gas, F2(g), to produce 2 moles of hydrogen fluoride gas, HF(g). This reaction is represented by the balanced chemical equation

H2(g)+F2(g) → 2HF(g)

Given:for H2(g): nH–H = 1 mol; DH–H = 432 kJ/mol; for F2(g): nF–F = 1 mol; DF–F = 154 kJ/mol;

Required:ΔHAnalysis: ∆H = Σn x D bonds broken –Σn x D bonds formedΔH = (nH-HDH-H + nF-F DF-F) – (nH-FDH-F 5.3

Solution:1 mol each of H–H and F–F bonds are brokenThe bonds formed are 2 mol of H–F bonds

∆H= (nH-HDH-H + nF-FDF-F) – nH-FDH-F

(1 mol x 432KJ) + (1 mol x 154 KJ) - (2 mol x 565 KJ mol mol mol

∆H = -544 KJ

The enthalpy change for the reaction of 1 mol hydrogen gas and 1 mol fluorine gas to ptoduce 2 mol. Hydrogen fluoride is -544 KJ

5.3

5.5: Standard enthalpy of formation

f

Standard enthalpy of formation (DH0) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm.

f

The standard enthalpy of formation of any element in its most stable form is zero.

DH0 (O2) = 0

DH0 (O3) = 142 kJ/molf

DH0 (C, graphite) = 0f

DH0 (C, diamond) = 1.90 kJ/molf

5.5

The standard enthalpy of reaction (DH0 ) is the enthalpy of a reaction carried out at 1 atm.

rxn

aA + bB cC + dD

DH0rxn dDH0 (D)fcDH0 (C)f= [ + ] - bDH0 (B)faDH0 (A)f[ + ]

DH0rxn nDH0 (products)f= S mDH0 (reactants)fS-

5.5

Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.

(Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)

Calculate the standard enthalpy of formation of CS2 (l) given that:C(graphite) + O2 (g) CO2 (g) DH0 = -393.5 kJrxn

S(rhombic) + O2 (g) SO2 (g) DH0 = -296.1 kJrxn

CS2(l) + 3O2 (g) CO2 (g) + 2SO2 (g) DH0 = -1072 kJrxn

1. Write the enthalpy of formation reaction for CS2

C(graphite) + 2S(rhombic) CS2 (l)

2. Add the given rxns so that the result is the desired rxn.

rxnC(graphite) + O2 (g) CO2 (g) DH0 = -393.5 kJ2S(rhombic) + 2O2 (g) 2SO2 (g) DH0 = -296.1x2 kJrxn

CO2(g) + 2SO2 (g) CS2 (l) + 3O2 (g) DH0 = +1072 kJrxn+

C(graphite) + 2S(rhombic) CS2 (l)

DH0 = -393.5 + (2x-296.1) + 1072 = 86.3 kJrxn5.5

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol.

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn nDH0 (products)f= S mDH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn = [ 12x–393.5 + 6x–285.8 ] – [ 2x49.04 ] = -6534.9 kJ

- 6534.9 kJ2 mol

= - 3267.44 kJ/mol C6H6

5.5

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is 49.04 kJ/mol.

2C6H6 (l) + 15O2 (g) 12CO2 (g) + 6H2O (l)

DH0rxn nDH0 (products)f= S mDH0 (reactants)fS-

DH0rxn 6DH0 (H2O)f12DH0 (CO2)f= [ + ] - 2DH0 (C6H6)f[ ]

DH0rxn = [ 12x–393.5 + 6x–285.8 ] – [ 2x49.04 ] = -6534.9 kJ

- 6534.9 kJ2 mol

= - 3267.44 kJ/mol C6H6

5.5

The enthalpy of solution (DHsoln) is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent.

DHsoln = Hsoln - Hcomponents

6.6

Which substance(s) could be used for melting ice?

Which substance(s) could be used for a cold pack?LiCl & CaCl2

NaCl, KCl, NH4Cl, NH4NO3